$\left[\begin{array}{l} a _{1} \\ a _{2}\end{array}\right]=\frac{1}{\sqrt{3}}\left[\begin{array}{cc}1 & -1 \\ 1 & k \end{array}\right]\left[\begin{array}{l} b _{1} \\ b _{2}\end{array}\right]$

$\left[\begin{array}{c}\sqrt{3} a_{1} \\ \sqrt{3} a_{2}\end{array}\right]=\left[\begin{array}{c}b_{1}-b_{2} \\ b_{1}+k b_{2}\end{array}\right]$

$b_{1}-b_{2}=\sqrt{3} a_{1} ....(1)$

$b _{1}+ kb _{2}=\sqrt{3} a _{2} ....(2)$

Given, $a_{1}^{2}+a_{2}^{2}=\frac{2}{3}\left(b_{1}^{2}+b_{2}^{2}\right)$

$(1)^{2}+(2)^{2}$

$\left(b_{1}+b_{2}\right)^{2}+\left(b_{1}+k b_{2}\right)^{2}=3\left(a_{1}^{2}+a_{2}^{2}\right)$

$a_{1}^{2}+a_{2}^{2}=\frac{2}{3} b_{1}^{2}+\frac{\left(1+k^{2}\right)}{3} b_{2}^{2}+\frac{2}{3} b_{1} b_{2}(k-1)$

Given, $a_{1}^{2}+a_{2}^{2}=\frac{2}{3} b_{1}^{2}+\frac{2}{3} b_{2}^{2}$

On comparing we get

$\frac{ k ^{2}+1}{3}=\frac{2}{3} \Rightarrow k ^{2}+1=2$

$\Rightarrow k =\pm 1 ....(3)$

And $\frac{2}{3}( k -1)=0 \Rightarrow k =1 ....(4)$

From both we get $k =1$